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Evaluate nth Roots and Use Rational Exponents
Chapter 6.1
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Nth Roots In an earlier chapter, we defined the square root of a number as follows If π₯ 2 =π, then π₯=Β± π When solving for a square, we can ask, βWhat number multiplies itself to produce a?β For example, if π₯ 2 =4, then x is the number that multiplies itself to produce 4 It turns out that there are actually two such numbers: 2 and β2, because 2 2 =4 and β2 2 =β2β
β2=4
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Nth Roots We can extend this idea further by asking of some number a, βWhat multiplies itself three times to produce a?β We can express this as π₯ 3 =π For example, what number multiplies itself three times to produce 8? This is the solution to the equation π₯ 3 =8 Since 2β
2β
2= 2 3 =8, then π₯=2; note that there is only one solution We can define the cube root of a number as If π₯ 3 =π, then π₯= 3 π
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Nth Roots We can define the cube root of a number as If π₯ 3 =π, then π₯= 3 π Notice that, unlike a square root, a cube root has only one answer Also unlike a square root, the cube root of any real number is a real number Recall that if π₯ 2 =β1, then x is not a real number; generally, if π₯ 2 equals any negative number, the result is imaginary But if π₯ 3 =β8, then π₯= 3 β8 =β2 because β2β
β2β
β2=β8
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Nth Roots Can we extend this further to 4th roots?
If π₯ 4 =π, then x is the number that multiplies itself four times to produce a For example, if π₯ 4 =81, then π₯=3 because 3β
3β
3β
3= 3β
3 β
3β
3 =9β
9=81 But x can also be β3 because β3β
β3 β
β3β
β3 =9β
9=81 So we define a fourth root as If π₯ 4 =π, then π₯=Β± 4 π
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Nth Roots So we define a fourth root as If π₯ 4 =π, then π₯=Β± 4 π
As will square roots, the fourth root of a negative number is imaginary For example, π₯ 4 =β16 has no real solution because there is no real number that multiplies itself 4 times to produce β16 We will only be concerned with real numbers in this chapter, so we will say that this has no real solution
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Nth Roots For the notation π π we would say βthe nth root of aβ
The whole number n is called the index of the radical and a is again called the radicand Only two roots have special names: square root and cube root Also, the index for a square root is not written; that is 2 π = π In defining nth roots, we must differentiate between even and odd roots
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n is an even whole number
Nth Roots Suppose that n is a whole number greater than 1, and that a is a real number such that π₯ π =π n is an even whole number n is an odd whole number If π<0, there are no real nth roots If π<0, then π₯= π π (1 solution, negative) If π=0, then π₯= π 0 =0 (1 solution) If π>0, then π₯=Β± π π (2 solutions) If π>0, then π₯= π π (1 solution, positive)
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Guided Practice Use the definition of nth roots and your calculator to find x. π₯ 5 =1024 π₯ 4 =16 π₯ 3 =β125 π₯ 6 =β1 π₯ 5 =2
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Guided Practice Use the definition of nth roots and your calculator to find x. π₯ 5 =1024; π₯= =4 because 4β
4β
4β
4β
4=1024 π₯ 4 =16; π₯= 4 16 =2 because 2β
2β
2β
2=16 π₯ 3 =β125; π₯= 3 β125 =β5 because β5β
β5β
β5=β125 π₯ 6 =β1; π₯= 6 β1 , but since the radicand is negative, there is no real solution π₯ 5 =2; π₯= 5 2 β because (1.1487)(1.1487)(1.1487)(1.1487)(1.1487)β2
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Rational Exponents Up to now we have only used integer exponents
Recall that π π =πβ
πβ
β¦β
π, where there are n factors of a Also, π βπ = 1 π π = 1 π β
1 π β
β¦β
1 π , where there are n factors of 1 π We can extend the meaning of an exponent to rational numbers (that is, to fractions) To do this we must recall one of the properties of exponents: The power property says that π π π = π ππ (multiply the exponents) Also recall that πβ
1 π =1 (inverse property of multiplication)
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Rational Exponents How can we interpret the equation π₯= 25 1 2 ?
Suppose that we raise both sides to the power 2: π₯ 2 = By the power property of exponents we get π₯ 2 = β
2 By the inverse property of multiplication, π₯ 2 = 25 1 =25 But by the definition of square root, π₯= 25 =5 (we donβt need both positive and negative here) Now, since π₯= and π₯= 25 , then 25 = =5
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Rational Exponents More generally, if π₯= π 1 π , then π₯ π = π 1 π π = π 1 π β
π = π 1 =π Since π₯ π =π, then by the definition of nth roots π₯= π π Now, π₯= π 1 π and π₯= π π , so we conclude that π 1 π = π π Rational exponents of the form 1 π are the same thing as nth roots
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Rational Exponents This means that we can rewrite nth roots of a number as the number raised to the 1 π power, and vice verse For example, 4 7 = , 3 9 = , and 2 = Remember that we donβt write the index for the square root Also, = 6 , = 5 3 , and = 4 10
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Guided Practice Rewrite the expression using rational exponent notation. 3 5 = 15 = 8 6 = π π =
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Guided Practice Rewrite the expression using rational exponent notation. 3 5 = 15 = 8 6 = π π = π 1 π
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Guided Practice Rewrite the expression using radical notation. 4 1 3 =
= = = π₯ 1 π =
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Guided Practice Rewrite the expression using radical notation.
= 3 4 = 5 11 = 3 π₯ 1 π = π π₯
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Rational Exponents With Denominator Other Than 1
Next we must interpret the meaning of π π as an exponent, where m is not 1 We can rewrite π π as πβ
1 π or as 1 π β
π (commutative property) Now, by the power property of exponents π 1 π π = π 1 π β
π = π π π But we also have that π 1 π = π π , so π 1 π π = π π π = π π π
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Rational Exponents With Denominator Other Than 1
Since multiplication is commutative, we have π π 1 π = π πβ
1 π = π π π But this can also be written π π 1 π = π π π = π π π In general, π π π = π π π = π π π Note that the denominator becomes the index and the numerator is an exponent in the radical or outside the radical (both are correct)
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Rational Exponents With Denominator Other Than 1
Example: rewrite the expression with rational exponents as a radical expression = = = 3 16 β2.520 = = = 4 32 β2.378 β = 5 β3 3 = 5 β3 3 = 5 β27 ββ1.933
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Guided Practice Rewrite the expression as a radical, then evaluate using a calculator. = = β = β =
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Guided Practice Rewrite the expression as a radical, then evaluate using a calculator. = = 3 2 =9 = = 2 3 =8 β = 3 β = β4 4 =256 β = 5 β9 3 ββ3.737
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Negative Rational Exponents
We defined negative exponents to mean π₯ βπ = π₯ βπ 1 = 1 π₯ π That is, you can move the numerator to the denominator and change the sign of the exponent We can understand negative rational exponents using this definition π₯ β π π = 1 π₯ π π = 1 π π₯ π = 1 π π₯ π , π₯β 0
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Negative Rational Exponents
Some examples 8 β = = = = 1 16 β64 β = 1 β = β = 1 β4 2 = 1 16 4 β = = = β0.574
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Guided Practice Rewrite the expression as a radical, then evaluate using a calculator. β30 β = 9 β = 243 β =
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Guided Practice Rewrite the expression as a radical, then evaluate using a calculator. β30 β = β = β0.104 9 β = = = β 243 β = = = 1 9
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Solving Radical Equations
We were able to use the square root property to solve equations We can also use the nth root definition to solve equations In general, if π₯ π =π, then π₯= π π Some examples 4 π₯ 5 =128βΉ π₯ 5 =32βΉπ₯= 5 32 =2 π₯β3 4 =21βΉπ₯β3=Β± 4 21 βΉ3Β± π₯= β5.141, π₯=3β 4 21 β0.859
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Guided Practice Solve the equation. Round the result to three decimal places when appropriate. π₯ 3 =64 1 2 π₯ 5 =512 π₯β2 3 =β14
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Guided Practice Solve the equation. Round the result to three decimal places when appropriate. π₯ 3 =64βΉπ₯= 3 64 =4 1 2 π₯ 5 =512βΉ π₯ 5 =1024βΉπ₯= =4 π₯β2 3 =β14βΉπ₯β2= 3 β14 βΉπ₯=2+ 3 β14 ββ0.410
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Exercise 6.1 Handout
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